The set of all x in −π2,π2 satisfying 4sinx−1<5 is given by
−π10,3π10
−π10,π5
−π10,3π5
−π5,3π10
We have −5<4sinx−1<5
⇒−(5−1)4<sinx<5+14
⇒−sinπ10<sinx<sin3π10
⇒−π10<x<3π10